A disk sliding with velocity u on a smooth horizontal plane strikes another identical disk kept at rest as shown in the figure. If the impact between the disks is perfectly elastic impact, find velocities of the disks after the impact.

Explanation

We first show velocity components along the t and the n-axis immediately before and after the impact. Angle that the line of impact makes with velocity u is `30^(@)`

Component along t-axis Components of momentum along the t-axis of each disk, considered separately, is conserved. Hence, t-component of velocities of each of the bodies remains unchanged.
`v_(At) = u_(At) = u/2` and `v_(Bt) = u_(Bt)=0 ` ...(i)
The component along the n-axis of the total momentum of the two bodies is conserved
`m_(B)v_(Bn) + m_(A)v_(An) = m_(B)u_(Bn) + m_(A)u_(An) rarr mv_(Bn) + mv_(An) = m xx 0 + m(usqrt3)/2`
`v_(Bn)+v_(An)=(usqrt3)/2` ...(ii)
Concept of coefficient of restitution e is applicable only for the n-component velocities.
`v_(Bn)-V_(An)=e(u_(An)-U_(Bn)) rarr " " v_(Bn)-V_(An)=(usqrt3)/2` ...(iii)
From equations (ii) and (iii), we have `v_(An) = 0` and `v_Bn=(usqrt3)/2` ...(iv)
From equations (i) and(iv) we can write velocities of both the disks.